Ebookscluborg-Electrical-Engineering-Principles-and-Applications-Fifth-Edition-Solutions-Manual

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1 CHAPTER 1 Exercises E1.1 Charge = Current × Time = (2 A) × (10 s) = 20 C E1.2 A ) 2cos(200 ) 200cos(200 0.01 0t) 0.01sin(20 ( ) ( ) ( t t dt d dt t dq t i = × = = = E1.3 Because i 2 has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element C . Because i 3 has a negative value, positive charge moves in the opposite direction to the reference. Thus positive charge moves upward in element E . E1.4 Energy = Charge × Voltage = (2 C) × (20 V) = 40 J Because v ab is positive, the positive terminal is a and the negative terminal is b . Thus the charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element. E1.5 i ab enters terminal a . Furthermore, v ab is positive at terminal a . Thus the current enters the positive reference, and we have the passive reference configuration. E1.6 (a) 2 20 ) ( ) ( ) ( t t i t v t p a a a = = J 6667 3 20 3 20 20 ) ( 3 10 0 3 10 0 10 0 2 = = = = = t t dt t dt t p w a a (b) Notice that the references are opposite to the passive sign convention. Thus we have: 200 20 ) ( ) ( ) ( = = t t i t v t p b b b J 1000 200 10 ) 200 20 ( ) ( 10 0 2 10 0 10 0 = = = = t t dt t dt t p w b b
2 E1.7 (a) Sum of currents leaving = Sum of currents entering i a = 1 + 3 = 4 A (b) 2 = 1 + 3 + i b i b = -2 A (c) 0 = 1 + i c + 4 + 3 i c = -8 A E1.8 Elements A and B are in series. Also, elements E , F , and G are in series. E1.9 Go clockwise around the loop consisting of elements A , B , and C : -3 - 5 + v c = 0 v c = 8 V Then go clockwise around the loop composed of elements C , D and E : - v c - (-10) + v e = 0 v e = -2 V E1.10 Elements E and F are in parallel; elements A and B are in series. E1.11 The resistance of a wire is given by A L R ρ = . Using 4 / 2 d A π = and substituting values, we have: 4 / ) 10 6 . 1 ( 10 12 . 1 6 . 9 2 3 6 × × × = π L L = 17.2 m E1.12 R V P 2 = = = 144 / 2 P V R A 833 . 0 144 / 120 / = = = R V I E1.13 R V P 2 = V 8 . 15 1000 25 . 0 = × = = PR V mA 8 . 15 1000 / 8 . 15 / = = = R V I E1.14 Using KCL at the top node of the circuit, we have i 1 = i 2 . Then, using KVL going clockwise, we have - v 1 - v 2 = 0; but v 1 = 25 V, so we have v 2 = -25 V. Next we have i 1 = i 2 = v 2 / R = -1 A. Finally, we have W 25 ) 1 ( ) 25 ( 2 2 = × = = i v P R and W. 25 ) 1 ( ) 25 ( 1 1 = × = = i v P s E1.15 At the top node we have i R = i s = 2A. By Ohm’s law we have v R = Ri R = 80 V. By KVL we have v s = v R = 80 V. Then p s = -v s i s = -160 W (the minus sign is due to the fact that the references for v s and i s are opposite to the passive sign configuration). Also we have W. 160 = = R R R i v P
3 Answers for Selected Problems P1.7* Electrons are moving in the reference direction (i.e., from a to b ). C 9 = Q P1.9* ( ) A 2 2 t t i + = P1.12* coulombs 2 = Q P1.14* (a) km 6 . 17 = h (b) m/s 9 . 587 = v (c) The energy density of the battery is J/kg 10 8 . 172 3 × which is about 0.384% of the energy density of gasoline. P1.17* coulombs 10 6 . 3 5 × = Q joules 10 536 . 4 Energy 6 × = P1.20* (a) 30 W absorbed (b) 30 W absorbed (c) 60 W supplied P1.22* C 50 = Q . Electrons move from b to a . P1.24* kWh 500 Energy = W 694.4 = P A 787 . 5 = I % 64 . 8 Reduction = P1.27* (a) P = 50 W taken from element A.

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